Question

2HCI (aq) + Pb (s) forms PbCl2 (aq)+ H2(g)

How many grams of lead chloride can be made from 100.0g of lead and 1.00L of 1.50M?

Answer 1

volume of HCl, V = 1.00 L

we have below equation to be used:

number of mol in HCl,

n = Molarity * Volume

= 1.5*1

= 1.5 mol

Molar mass of Pb = 207.2 g/mol

mass of Pb = 100.0 g

we have below equation to be used:

number of mol of Pb,

n = mass of Pb/molar mass of Pb

=(100.0 g)/(207.2 g/mol)

= 0.4826 mol

we have the Balanced chemical equation as:

2 HCl + Pb ---> PbCl2 + H2

2 mol of HCl reacts with 1 mol of Pb

for 1.5 mol of HCl, 0.75 mol of Pb is required

But we have 0.4826 mol of Pb

so, Pb is limiting reagent

we will use Pb in further calculation

Molar mass of PbCl2 = 1*MM(Pb) + 2*MM(Cl)

= 1*207.2 + 2*35.45

= 278.1 g/mol

From balanced chemical reaction, we see that

when 1 mol of Pb reacts, 1 mol of PbCl2 is formed

mol of PbCl2 formed = (1/1)* moles of Pb

= (1/1)*0.4826

= 0.4826 mol

we have below equation to be used:

mass of PbCl2 = number of mol * molar mass

= 0.4826*2.781*10^2

= 134 g

Answer: 134 g